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Can a meson be in a pure $b \overline{b}$, $r \overline{r}$, $g \overline{g}$ state or does it have to be in the $\frac{1}{\sqrt{3}}\left(b \overline{b}+r \overline{r}+g \overline{g}\right)$ state?

Why?

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/91609/2451 $\endgroup$ – Qmechanic Dec 23 '20 at 22:35
  • $\begingroup$ Is $b \overline{b}$ a color singlet? I.e. invariant under a color rotation? $\endgroup$ – Cosmas Zachos Dec 23 '20 at 23:00
  • $\begingroup$ Does it have to be? $\endgroup$ – Stijn Boshoven Dec 23 '20 at 23:01
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    $\begingroup$ All hadrons are color singlets... Color is a gauge symmetry ~ not observable. $\endgroup$ – Cosmas Zachos Dec 24 '20 at 1:43
  • $\begingroup$ @CosmasZachos How do we know it is not an observable? Cannot the alternative explanation that we just don't see the difference between the different colours be given to explain this? Or maybe this is why I've been seeing people talking about Occam's razor in the context of this. If there is some difference between the particles that cannot be physically measured, we should assume them all to be the same and only consider singlet states. $\endgroup$ – Stijn Boshoven Dec 24 '20 at 18:16
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Another way of saying the same thing, is that if/when a meson is in a $b \overline b$ state it can annihilate through gluons and form a $r \overline r$ state with the same quark flavours, and likewise a $g \overline g$ state. The 3 states all mix into each other: you can't have a $b \overline b$ meson because it won't stay a $b \overline b$ meson. The eigenstates of the mixing (i.e. the states which will stay the same over time) are $(b \overline b + g \overline g + r \overline r)/\sqrt 3$, $(r \overline r - g \overline g)/\sqrt 2$ and $(r \overline r + g \overline g - 2 b \overline b)/\sqrt 6$. Then you use the fact that the first of these has colour zero which is allowed and the second (degenerate) two have total colour 1 and are forbidden.

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  • $\begingroup$ So even if you have a $b\bar{b}$ state, it will very quickly go into a superposition of all three colours evenly, which would be the singlet state? Because of the timescales it then doesn't really make sense to ever consider anything but the singlet state. I don't understand why those other two states you mentioned should stay the same or why having a total colour is forbidden. Exchange of gluons between two quarks of opposite colour charge is attractive. The idea that colour is not an observable doesn't convince me that singlet states are the only real ones. But maybe this is Occams razor. $\endgroup$ – Stijn Boshoven Dec 24 '20 at 18:08
  • $\begingroup$ @StijnBoshoven so essentially your follow-up question is, "If hadrons are colour singlet states, why should we even consider any coloured state"? Or are you wondering about how the gauge symmetry implies that hadrons must be colourless? $\endgroup$ – Nihar Karve Dec 25 '20 at 15:54
  • $\begingroup$ @NiharKarve indeed I'm wondering how the gauge symmetry implies this. Just the fact that the singlet is unchanged under SU(3) transformation doesn't explain it as far as I can currently see. $\endgroup$ – Stijn Boshoven Dec 28 '20 at 15:41
  • $\begingroup$ @StijnBoshoven I have updated my answer $\endgroup$ – Nihar Karve Dec 28 '20 at 16:31
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Due to colour confinement, observed free particles (hadrons) must be "colourless" or "white", i.e. a colour singlet. A necessary (but not sufficient) condition for a colour singlet is that it is invariant under the $\text{SU}(3)$ colour gauge symmetry, which automatically rules out "pure" $r\bar{r}$, $b\bar{b}$ and $g\bar{g}$ mesons by inspection - such pure states would mix under an $\text{SU}(3)$ transform and so would not be colourless.

Since mesons are a bound state of one quark and one anti-quark, you can decompose the tensor product of the fundamental and anti-fundamental representations of the colour space: $\mathbf{3 \otimes \bar{3}}= \mathbf{8\oplus1}$, which breaks down the nonet into a colour octet and a colour(less) singlet - this singlet is then identified with $\frac{1}{\sqrt{3}}\left(r\bar{r} + b\bar{b} +g\bar{g}\right)$. This is analogous to identifying the flavour singlet with the eta meson in the approximate $\text{SU(3)}_{\rm flavour}$ symmetry: see Qmechanic's answer here. A visual depiction of this is: [Source: Mark Thomson's QCD Lecture Slides]

enter image description here

[edit in response to follow-up question]:

The reason why colour confinement should exist at all to force observable bound states to be colour singlets has no rigorous underpinnings in our current model of QCD, or any non-abelian gauge theory for that matter. Colour confinement, being a low-energy phenomenon, is resistant to the tools of perturbative QCD, and can only be demonstrated somewhat heuristically in effective field theories that operate at those energy scales, such as chiral perturbation theory (in addition to other heuristics such as "$\text{SU}(3)$ is a colour gauge symmetry, so rotations of the bound states must act trivially" - this has very firm grounding, but may seem hacky at first sight). In fact, demonstrating this axiomatically is equivalent to proving one of the Millenium Prize Problems: the Yang-Mills and Mass Gap problem, which, as such, will net you $1 million from the Clay Math Institute (see here for the relation between the two problems as well as further exposition on the rigour behind colour confinement)

In response to the "if hadrons are 'colourless' anyway, why even consider separate colours?" line of thought, the difference in the interaction between two red-antired mesons vs. between a red-antired and a blue-antiblue meson is measurable, amongst many other testable results that have been confirmed. It might be worth reading up on how the idea of colour charge popped up in the first place, c.f. the $\Omega^-$ and $\Delta^{++}$ crises.

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